Suppose 60% of a population of part-time college students is female. In a simulation, we collected thousands of random samples from a population of part-time college students. What is the mean of the sample proportions?
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If the population mean is 60%,  you would expect the sample means to be 60% in view of the large number of samples.
by Top Rated User (1.1m points)

About 28%. There is about a 28% chance that a random sample of 100 adults will have a sample proportion with more than 0.05 error in either direction. In other words, there is about a 28% chance that sample proportions fall below 0.25 or above 0.35 if the true population proportion is 0.30.

Here is how we got this probability.

Check normality conditions:

Conditions are met. In a sample of 100, we expect 30% successes and 70% failures. 100(0.30) = 30 and 100(0.70) = 70. So we can use a normal model.

Find the Z-score:

We want the error to be more than 5%. So the sample proportion could be less than 0.25 or greater than 0.35. It does not matter which sample proportion we use to find the Z-score because of the symmetry in the distribution. We arbitrarily chose 0.35. You could also have used 0.25.

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Because of the symmetry in the distribution, we know that the Z-score for a sample proportion of 0.25 is -1.09.

 

 

We want the probability described by the two tails. The probability for one tail is 0.1379, or about 0.14. So the probability for both tails is about 2 x 0.14 = 0.28.

Conclusion: If it is true that 30% of smokers started smoking before the age of 16, then there is a 28% chance that the percentage from a random sample of 100 smokers is off by more than 5%. In other words, there is a 28% chance that the sample proportion is either less than 0.25 or greater than 0.35.

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